In Young's double slit experiment, two slits | vedclass

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(B) In Young's double slit experiment, the path difference $\Delta x$ between the light waves reaching point $P$ from slits $A_1$ and $A_2$ is given b

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$\sqrt{\lambda D}$

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(B) In Young's double slit experiment, the path difference $\Delta x$ between the light waves reaching point $P$ from slits $A_1$ and $A_2$ is given by $\Delta x = A_2P - A_1P$.Given that $A_1P$ is perpendicular to $A_1A_2$, we have a right-angled triangle $\Delta A_1A_2P$ where $A_1A_2 = a$ and $A_1P = D$.Using the Pythagorean theorem, $A_2P = \sqrt{A_1A_2^2 + A_1P^2} = \sqrt{a^2 + D^2}$.Thus, the path difference is $\Delta x = \sqrt{a^2 + D^2} - D$.For the first minimum (destructive interference), the path difference must be equal to $\frac{\lambda}{2}$.So, $\sqrt{a^2 + D^2} - D = \frac{\lambda}{2}$.$\sqrt{a^2 + D^2} = D + \frac{\lambda}{2}$.Squaring both sides, $a^2 + D^2 = (D + \frac{\lambda}{2})^2 = D^2 + \lambda D + \frac{\lambda^2}{4}$.$a^2 = \lambda D + \frac{\lambda^2}{4}$.Since $\lambda$ is very small compared to $D$, $\frac{\lambda^2}{4}$ can be neglected.Therefore, $a^2 \approx \lambda D$, which gives $a = \sqrt{\lambda D}$.

In Young's double slit experiment, two slits are illuminated with light of wavelength $\lambda$. The line joining $A_1P$ is perpendicular to $A_1A_2$. If the first minimum is detected at $P$, the value of slit separation '$a$' will be (where $D$ is the distance between the source and the screen):

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